Centre of mass MCQ Quiz in मराठी - Objective Question with Answer for Centre of mass - मोफत PDF डाउनलोड करा

CONCEPT:

A group of particles that undergoes a particular type of motion together is called a system of particles.

A particular point where the whole of the mass of the system of particles appeared to be concentrated is called the center of mass.

Let us consider two-particle systems having masses m₁ and m₂ located on X-axis at x₁ and x₂ respectively.

The Centre of the mass of the two particles is at point c which is given by:

\(X = \frac{{{m_1}{x_1} + {m_2}{x_2}}}{{{m_1} + {m_2}}}\;\)

If m₁ = m₂ = m

\(X = \frac{{m\;{x_1} + m{x_2}}}{{m + m}} = \frac{{{x_1} + {x_2}}}{2}\)

If we have n particles having masses m₁, m₂, m₃, …….…m_n on a straight line then Centre of mass is given by:

\(X = \frac{{{m_1}{x_1} + \;{m_2}{x_2} + {m_3}{x_3} + \; \ldots \ldots .{m_n}{x_n}}}{{{m_1} + {m_2} + \;{m_3} + \; \ldots \ldots \ldots .{m_n}}} = \;\frac{{\mathop \sum \nolimits_{i = 1}^n {m_i}{x_i}}}{{\mathop \sum \nolimits_{i = 1}^n {m_i}}}\)

Now suppose we have three particles of masses m₁, m₂ and m₃ not on a straight line but in space at (x₁, y₁, z₁), (x₂, y₂, z₂) and (x₃, y₃, z₃). Then the Centre of the mass of the system is located at (X, Y, Z) which is given by:

\(X - coordinate\;\left( X \right) = \frac{{{m_1}{x_1} + \;{m_2}{x_2} + {m_3}{x_3}}}{{{m_1} + {m_2} + \;{m_3}}}\)

\(Y - coordinate\;\left( Y \right) = \frac{{{m_1}{y_1} + \;{m_2}{y_2} + {m_3}{y_3}}}{{{m_1} + {m_2} + \;{m_3}}}\)

\(Z - coordinate\;\left( Z \right) = \frac{{{m_1}{z_1} + \;{m_2}{z_2} + {m_3}{z_3}}}{{{m_1} + {m_2} + \;{m_3}}}\)

CALCULATION:

Given that,

Mass of particle A, m₁ = 0.5m

Mass of particle A, m₂ = m

Mass of particle A, m₃ = 1.5m

the distance of particle A from its origin, x₁ = 3x

the distance of particle B from its origin, x₂ = 2x

the distance of particle C from its origin, x₃ = x

Also, from the above-given explanation, we can see that center of mass for the two-particle system can be expressed as

\(X = \frac{{{m_1}{x_1} + \;{m_2}{x_2} + {m_3}{x_3} + \; \ldots \ldots .{m_n}{x_n}}}{{{m_1} + {m_2} + \;{m_3} + \; \ldots \ldots \ldots .{m_n}}} = \;\frac{{\mathop \sum \nolimits_{i = 1}^n {m_i}{x_i}}}{{\mathop \sum \nolimits_{i = 1}^n {m_i}}}\)

Now in this case for the three-particle system, it can be modified as

C.M along X-coordinate \(\left( {{x_{cm}}} \right) = \frac{{{m_1}{x_1} + \;{m_2}{x_2} + {m_3}{x_3}}}{{{m_1} + {m_2} + \;{m_3}}}\) 

\(\therefore {x_{cm}} = \frac{{0.5m \times 3x + m \times 2x + 1.5m \times x}}{{0.5m + m + 1.5m}} = \frac{{5mx}}{{3m}} = \frac{5}{3}x\)

CONCEPT:

→The center of mass is defined as the point in which the whole mass of the body or all the masses appeared to be concentrated we can say that center of mass is defined as the relative position of mass with respect to the object and it is written as;

\(x_{cm} = \frac{m_1x_1+m_2x_2}{m_1+m_2}\)

Here we have m as the mass and x as the distance.

EXPLANATION:

Let us see the given options one by one.

→In option 1, it is given that pencil, the pencil is having the shape of a cylinder and the center of the mass lies inside the cylindrical shape which is not outside the body.

Therefore, option 1) is not correct.

→In option 2, A shotput, shotput is having the shape of a solid sphere, and the center of the mass lies inside the cylindrical shape which is not outside the body.

Therefore, option 1) is not correct.

→In option 3, A dice, shotput is having the shape of a solid cube, and the center of the mass lies inside the cube which is not outside the body.

Therefore, option 1) is not correct.

→In option 4, A bangle which is having the shape of a ring in which the center of mass lies inside the ring which lies outside the body.

Therefore option 4) is the correct answer.

Hence option 4) is the correct answer.

Concept:

Centre of mass:

The centre of mass of a body or system of a particle is defined as, a point at which the whole of the mass of the body or all the masses of a system of particles appeared to be concentrated.

The motion of the centre of mass:

Let there be n particles of masses m1, m2,...,mn.

If all the masses are moving then,

⇒ Mv = m1v1 + m2v2 + ... + mnvn

⇒ Ma = m1a1 + m2a2 + ... + mnan

⇒ \(M\vec{a}=\vec{F_1}+\vec{F_2}+...+\vec{F_n}\)

⇒ M = m1 + m2 + ... + mn

Thus, the total mass of a system of particles times the acceleration of its centre of mass is the vector sum of all the forces acting on the system of particles.

Thus the internal forces contribute nothing to the motion of the centre of mass.

Explanation:

The body explodes somewhere during the flight and splits into a number of particles.

When the body explodes only internal forces act on the body and we know that the internal forces contribute nothing to the motion of the centre of mass.

Therefore the centre of mass will follow the same parabolic path after the explosion.

Concept:

• Linear mass density is defined as mass per unit length.
• Centre of mass is an Imaginary point where the whole mass of the system can be assumed to be concentrated.
• Centre of mass for the system of multiple masses m₁,m₂,m₃ up to n masses m_n is given as.

⇒ x_cm = \(\frac{m_1x_1+m_2x_2+m_3x_3+...+m_nx_n}{m_1+m_2+m_3+...+m_n}\)

Explanation:

• Given Data
• Linear mass density is given as λ = Ax, where A is constant (∴ λ = \(\frac{dm}{dx}\))
• Integration the masses from origin x = 0 to the length x = L,

⇒ x_cm = \(\frac{\int_0^Lxdm}{\int_0^Ldm}\) = \(\frac{\int_0^Lλ Xdx}{\int_0^Lλ dx}\)

⇒ x_cm = \(\frac{\int_0^L(Ax) xdx}{\int_0^LAx dx}\) = \(\frac{\int_0^LAx^2dx}{\int_0^LAx dx}\)

⇒ x_cm =  \(\frac{[\frac{x^3}{3}]^L_0}{[\frac{x^2}{2}]_0^L}\)

⇒ x_cm = \([\frac{2x^3}{3x^2}]_0^L\)

⇒ x_cm = \(\frac{2}{3}\)L

The correct answer is option 2) i.e. To lower their centre of mass.

CONCEPT:

• Centre of mass: The point on a body where the whole of its mass is assumed to be concentrated is known as the centre of mass of the body.
• The lower the centre of mass, the more stable is the body.

EXPLANATION:

• Circus artist uses long poles so as to lower their centre of mass and thus provide themselves with stability and better performance.
• A lower centre of mass reduces the possibility of overturning the body and thus helps in maintaining stability.

Additional Information

• A higher centre of mass increases the possibility of overturning because the body reduces stability and a small force can tip it off or overturn it. 

Concept:

Centre of Gravity:

Centre of gravity of a body is the point through which the whole weight of the body acts. A body is having only one centre of gravity for all positions of the body. It is represented by C.G or simply G.

Explanation:

Considering the strip parallel to X-axis

Area of strip, dA = 2x . dy

The distance of the C.G. of this area from the x-axis is y

∴ Moment of this area about the x-axis

= y. dA

= y. 2xdy

= 2xy dy  ......................(i)

But, we know x² + y² = R²

∴ x² = R² - y²

or, x = \( \sqrt{ R^2 - y^2} \)

Substituting the above value of x in equation (i), we get

Moment of area dA about the x-axis,

= 2 \( \sqrt{ R^2 - y^2} \) . y . dy

The moment of total area A about the x-axis will be obtained by integrating the above equation from 0 to R.

∴ Moment of area A about x-axis

= \(\rm \int_0^R2\sqrt{R^2-y^2}.y\ dy\)

(∴ y varies from 0 to R)

= \(-\rm \int_0^R2\sqrt{R^2-y^2}.(-2y)\ dy=-\left[\frac{(R^2-y^2)^{3/2}}{3/2}\right]_0^R\)

\(\rm =-\frac{2}{3}[0-R^2]=\frac{2R^3}{3}\)      ..........(ii)

Also the moment of total area A about x-axis = A × y̅

Where

A = Total area of semi-circle = \(\rm \frac{\pi R^2}{2}\)

y̅ = Distance of C.G of area A from x-axis

∴ Moment of total A about x-axis = \(\rm \frac{\pi R^2}{2}\) × y̅      ..........(iii)

 Equating the two values given by equations (ii) and (iii),

\(\rm \frac{\pi R^2}{2}\) × y̅ = \(\rm\frac{2R^3}{3}\)

or y̅ = \(\frac{2R^3}{3}\) x \(\frac{2}{\pi R^2}\)

or y̅ = \(\frac{4R}{3\pi }\)

Calculation:

Given that,

R = \(\frac{D}{2}\) = \(\frac{1.32}{2}\) = 0.66

y̅ = \(\frac{4R}{3\pi }\)

y̅  = \(\frac{4 X 0.66}{3 X \frac{22}{7} }\)

y̅  = 0.28 m

So the correct answer is Option 3

Concept Used:-

For the two bodies of mass m₁ and mass m₂ the center of along the y-axis lies at distance Y which can be given as,

\(Y=\frac{m_1 y_1+m_2 y_2}{m_1+m_2}\)

Where y₁ and y₂ is the place where the center of gravities of body 1 and body 2 lies respectively along the y-axis.

Explanation:-

Let's suppose the height of the cylinder is h and the radius of the base of the cylinder is r. Also, suppose that ρ indicates the densities of solids.

Since the base of the cylinder and hemisphere is the same. So, the radius of the hemisphere base is also r. 

Now, we know that,

Volume of cylinder V₁ = πr2h

Mass of cylinder m₁ = ρ(πr2h)

Volume of hemisphere V2 = 2/3(πr3

Mass of hemisphere m2 = ρ[2/3(πr3]

Since we know that the center of gravity of the cylinder is in the middle (at a distance of h/2). So,

⇒ y1 = h/2 

Also, center of gravity (COG) of a hemisphere lies on a central radius from the plane base is at a distance of (3r/8).

So,

⇒ y2 = 3r/8 

Now let's consider the figure below,

Now, when the combined center of gravity lie at the center of the base, then the value of this position along y-axis will be zero. So, from the above formula,

\(\Rightarrow Y=\frac{m_1 y_1+m_2 y_2}{m_1+m_2}\\ \Rightarrow 0=\frac{\mathrm{p} \times \pi r^2 h \times \frac{h}{2}+p \times \frac{2}{3} \pi r^3\left(-\frac{3 r}{8}\right)}{p \times \pi r^2 h+p \times \frac{2}{3} \pi r^3} \\ \Rightarrow p \times \pi r^2\left[\frac{h^2}{2}-2 r \times \frac{3 r}{8}\right]=0 \\ \Rightarrow \frac{h^2}{2}-\frac{ r^2}{4}=0 \\ \Rightarrow \frac{h^2}{2}=\frac{ r^2}{4} \\ \Rightarrow \dfrac{h}{r}=\sqrt{\dfrac{1}{2}} \)

So, the ratio of the height of the cylinder to the radius of the base will be 1 ∶ √2.

Hence, the correct option is 2.

The correct answer is option 3) i.e. Either inside or outside the body.

CONCEPT:

• Centre of mass: The point where the whole mass of a body is assumed to be concentrated.
• The Centre of mass may lie either inside or outside the body.

EXPLANATION:

• The position of the centre of mass of an object depends on the distribution of mass over the body.
• For a solid sphere, the centre of mass is at its centre i.e. inside the sphere.
• For a solid ring, the centre of mass is at its geometric centre i.e. outside the ring.

CONCEPT:

• Center of mass (RCM): The center of mass of the body is a point where the whole mass of the body may be assumed to be concentrated for describing its translatory motion.
  • The center of mass of a system of particles is that single point that moves in the same way in which a single particle having the total mass of the system and acted upon by the same external force would move.
• Mathematically center of mass for n – particle is written as

\({\vec R_{CM}} = \frac{{{m_1}{{\vec r}_1} + {m_2}{{\vec r}_2} + {m_3}{{\vec r}_3} - - - - - - - {m_n}{{\vec r}_n}}}{{{m_1} + {m_2} + {m_3} - - - - - - - {m_n}}}\)

EXPLANATION:

• Since a uniform rod is a symmetrical body that has no variation in its mass distribution throughout the body. So the center of mass will be at the middle point of the rod. Hence option 1 is correct.

Concept:

• Center of mass (CM): The center of mass of the body is a point where the whole mass of the body may be assumed to be concentrated for describing its translatory motion.
  • The center of mass of a system of particles is that single point that moves in the same way in which a single particle having the total mass of the system and acted upon by the same external force would move.
• Mathematically center of mass for n – particle is written as

\(CM = \frac{{{m_1}{{\vec r}_1} + {m_2}{{\vec r}_2} + {m_3}{{\vec r}_3} - - {m_n}{{\vec r}_n}}}{{{m_1} + {m_2} + {m_3} - - {m_n}}} \)

• The concept of center of mass is useful in analyzing the complicated motion of the system of objects, particularly When two and more objects collide or an object explodes into fragments.

Explanation:

• When an explosive shell travelling in a parabolic path under the effect of gravity explodes, the center of mass of the fragments will move along the original parabolic path.
• We know that explosive shell travelling in a parabolic path under gravity so,

\( Before\;\exp losion= \frac{{\overrightarrow {{F_{ext}}} }}{M} = \frac{{mg}}{M} = g\)

\( After\;\exp losion\; = \frac{{{m_1}g + {m_2}g + ....}}{{{m_1} + {m_2} + ....}} = g \)

• The internal forces have no effect on the trajectory of the center of mass, and the forces due to explosion are the internal forces.

So the center of mass will follow the same parabolic path even after the explosion.